Nonlinear dynamics & chaos BECS-114.7151
Phase portraits Focus: nonlinear systems in two dimensions General form of a vector field on the phase plane: Vector notation:
Phase portraits Solution x(t) describes a trajectory on the phase plane The whole plane is filled with (non-intersecting) trajectories For nonlinear systems, there is no hope to find trajectories analytically Our approach: determining the qualitative behavior of the solutions!
Phase portraits Huge variety of possible phase portraits Salient features: 1) Fixed points (A, B, C): equilibria of the system 2) Closed orbits (D): periodic solutions 3) Arrangement of trajectories near fixed points and closed orbits 4) Stability or instability of the fixed points and closed orbits
Numerical computation of Runge-Kutta method phase portraits A stepsize Δt=0.1 usually provides sufficient accuracy
Example I Fixed points Stability Exponentially growing solutions: the fixed point is unstable
Example I Phase portrait: nullclines The nullclines are the curves where On the nullclines the flow is either purely horizontal or purely vertical
Example I The fixed point is a nonlinear version of a saddle point
Existence, uniqueness and topological consequences Are we sure that there is a solution? Existence and Uniqueness Theorem: let f be continuous and all its partial derivatives be continuous for x in some open connected set. Then for the initial value problem has a solution x(t) on some time interval (-τ, τ) about t=0, and the solution is unique
Existence, uniqueness and topological consequences Corollary: different trajectories never intersect! If two trajectories did intersect there would be two solutions starting from the same point (the crossing point)
Existence, uniqueness and topological consequences Consequence: any trajectory starting from inside a closed orbit will be trapped inside it forever! What will happen in the limit of t ->? The trajectory will either converge to a fixed point or to a periodic orbit!
Fixed points and linearization Fixed point (x *, y * ) Does the disturbance from the fixed point grow or decay?
Fixed points and linearization Likewise:
Fixed points and linearization Jacobian matrix at the fixed point (x *, y * ), multivariate analog of the derivative f (x*) for 1-dimensional systems If we neglect second-order terms and higher, we obtain the linearized system
Fixed points and linearization Dynamics can be analyzed with the methods used for linear systems Effect of small nonlinear terms If the fixed point is not one of the borderline cases (centers, degenerate nodes, stars, non-isolated fixed points) the predicted type of the linearized system is the correct one
Example I Fixed points: (0,0), (1,0), (-1,0) (0,0) -> (±1,0) -> Stable node Saddle points
Example I 1) x and y equations are uncoupled 2) y-direction: trajectories decay exponentially to y=0 3) x-direction: trajectories are attracted to x=0 and repelled by x= ± 1 4) Vertical lines x=0 and x = ± 1 are invariant: if the particle is there it will always stay there 5) The axis y=0 is invariant too 6 4 2 0 x -2-4 -6-2 -1.5-1 -0.5 0 0.5 1 1.5 2
Example I Phase portrait is symmetric with respect to the x and the y axes
Example II (0,0) is a fixed point -> linearization τ = 0, Δ=1 > 0 -> the fixed point is a center (borderline case) Is that so? Let us switch to polar coordinates
Example II
Example II
Example II Radial and angular motions are independent The fixed point is a spiral (stable for a < 0, unstable for a > 0) Centers are delicate: one needs that the orbit closes perfectly after one cycle, the slightest perturbation turns it into a spiral
Fixed points and linearization Stars and degenerate nodes can be altered by small nonlinearities, but their stability does not change! Stars and degenerate nodes stay in fact well within regions of stability or instability: small perturbations will leave them in those areas.
Fixed points and linearization Robust cases 1) Repellers (or sources): both eigenvalues have positive real part 2) Attractors (or sinks): both eigenvalues have negative real part 3) Saddles: one eigenvalue is positive, the other is negative Marginal cases 1) Centers: both eigenvalues are purely imaginary 2) Higher-order and non-isolated fixed points: at least one eigenvalue is zero Marginal cases are those where at least one eigenvalue satisfies Re(λ) = 0
Fixed points and linearization If Re(λ) 0 for both eigenvalues, the fixed point is called hyperbolic: in this case its type is predicted by the linearization Same thing in higher-order systems Hartman-Grobman Theorem: the local phase portrait near a hyperbolic fixed point is topologically equivalent to the phase portrait of the linearization
Fixed points and linearization Intuition: two phase portraits are equivalent if one is a distorted version of the other What type of distortion? Bending, warping, but not ripping Consequences: closed orbits stay closed, trajectories connecting saddle points must not be broken, etc. A phase portrait is structurally stable if its topology cannot be changed by an arbitrarily small perturbation to the vector field Examples: the phase portrait of a saddle point is structurally stable, that of a center is not, as a small perturbation converts the center into a spiral
Rabbits versus Sheep
Rabbits versus Sheep Lotka-Volterra model of competition between two species Rabbits and sheep competing for the same limited resource (e.g. grass): no predators, seasonal effects, etc. 1) Each species would grow to its carrying capacity in the absence of the other -> logistic growth 2) When rabbits and sheep encounter each other, trouble starts: sheep kick rabbits away -> conflicts occur at a rate proportional to the size of each population, reducing the growth rate for each species 3) Rabbits reproduce faster but they are more penalized by conflicts
Rabbits versus Sheep x(t) 0 -> population of rabbits y(t) 0 -> population of sheep Fixed points (0,0), (0,2), (3,0), (1,1)
Rabbits versus Sheep (0,0) Eigenvalues are λ= 2, 3 -> the origin is an unstable node Trajectories near a node are tangential to the slower eigendirection (here the y-axis, for which λ = 2 < 3)
Rabbits versus Sheep (0,2) Eigenvalues are λ= -1, -2 -> (0,2) is a stable node Trajectories near a node are tangential to the slower eigendirection [here v = (1, -2), for which λ = -1 -> -1 < - 2 ]
Rabbits versus Sheep (3,0) Eigenvalues are λ= -3, -1 -> (3,0) is a stable node Trajectories near a node are tangential to the slower eigendirection [here v = (3, -1), for which λ = -1 -> -1 < - 3 )]
Rabbits versus Sheep (1,1) Eigenvalues are λ= -> (1,1) is a saddle point
Rabbits versus Sheep The xy axes contain horizontal/vertical trajectories
Rabbits versus Sheep Biological interpretation: one species drives the other to extinction!
Rabbits versus Sheep Principle of competitive exclusion: two species competing for the same limited resource typically cannot coexist Basin of attraction of an attracting fixed point: set of initial conditions leading to that fixed point as t ->
Conservative systems Equation of motion of a mass subject to a nonlinear force F(x): F(x) has no dependence on the velocity or time -> no damping or friction, no time-dependent driving force Energy is conserved V(x) is the potential energy
Conservative systems Systems with a conserved quantity are called conservative General definition: given a system a conserved quantity is a real-valued continuous function E(x) that is constant on trajectories (de/dt=0), but nonconstant on every open set
Example I A conservative system cannot have any attracting fixed points If there were a fixed point x *, all points in its basin of attraction are bound to have the same energy E(x * ) (since energy would be constant on all trajectories leading to x * ), so there would be an open set with constant energy What kind of fixed points can there be?
Example II Particle of mass m=1 moving in a double-well potential Fixed points: (0, 0), (±1,0)
Example II (0, 0) -> Δ = -1 <0 -> saddle point! (±1, 0) -> τ=0, Δ = 2 -> centers! Question: will the centers survive the nonlinearity? Answer: yes, because of energy conservation!
Example II
Example II 1) Near the centers there are small periodic orbits 2) There are also large periodic orbits encircling all fixed points 3) Solutions are periodic except for equilibria (fixed points) and the homoclinic orbits, which approach the origin when t -> ±
Example II 1) Neutrally stable equilibria correspond to the particle at rest at the bottom of either one of the wells 2) Small closed orbits -> small oscillations about equilibria 3) Large closed orbits -> oscillations taking the particle back and forth over the hump 4) Saddle point? Homoclinic orbits?
Example II Sketch the graph of the energy function 1) Local minima of E project down to centers in the phase plane 2) Contours of slightly higher energy -> small closed orbits 3) Cut at E-value of local maximum -> homoclinic orbits 4) Cuts at higher E-values -> large periodic orbits
Nonlinear centers Theorem (nonlinear centers for conservative systems): Consider the system, where and f is continuously differentiable. Suppose there exists a conserved quantity E(x) and an isolated fixed point x *. If x * is a local minimum of E, then all trajectories sufficiently close to x * are closed. Ideas behind the proof: 1) Since E is constant on trajectories, each trajectory is contained in some contour of E. 2) Near a local maximum (or minimum), contours are closed 3) The orbit is periodic, i.e. it does not stop at some point of the contour because x * is isolated, so there are no other fixed points on its close proximity
Reversible systems Many mechanical systems have time-reversal symmetry, i.e. if one watches the motion backwards no anomaly is seen, the reversed motion is a possible future trajectory of the system (Ex. Pendulum) Any system is symmetric under time reversal! The acceleration does not change, the velocity changes sign!
Reversible systems Consequence: if [x(t), y(t)] is a solution, also [x(-t), -y(-t)] is a solution!
Any system Reversible systems such that f(x,-y) = -f(x,y) and g(x,-y)=g(x,y) is reversible! Reversible systems are different from conservative systems, but they share some properties Theorem (nonlinear centers for reversible systems): Suppose the origin x * = 0 is a linear center of a reversible system. Then, sufficiently close to the origin, all orbits are closed.
Reversible systems Ideas behind the proof: 1) Let us take a trajectory starting on the positive x-axis near the origin 2) Because of the influence of the linear center (if the system is close enough to it), the trajectory will bend and intersect the negative x-axis
Reversible systems Ideas behind the proof: 3) By using reversibility we can reflect the trajectory above the x-axis, obtaining a twin trajectory (we know that it is a solution of the equation of motion, it must be the only one) 4) The two trajectories form a closed orbit, as desired
Example I The system is reversible and the origin (0, 0) is a fixed point, which kind? Jacobian at the origin: τ = 0, Δ = 1 -> linear center -> nonlinear center (theorem) Other fixed points are (-1, 1) and (-1, -1) Δ = -2 < 0 -> saddle points
Example I The twin saddle points are joined by a pair of trajectories (heteroclinic orbits or saddle connections) Homoclinic and heteroclinic orbits are more common in conservative and reversible systems
Example II Show that there is a homoclinic orbit in the half-plane x 0 The origin is a saddle point, since the Jacobian is which has Δ = -1 < 0 The eigenvectors are v 1 = (1, 1) and v 2 = (1, -1) The unstable manifold leaves the origin following v 1 = (1, 1)
Example II 1) Initially we are in the first quadrant (x > 0 and y > 0) 2) Velocity in the x direction is positive, in the y direction it is positive until the system passes x = 1 3) For x > 1 the velocity in the y direction becomes negative and the particle ends up hitting the x-axis 4) By reversibility there must be a twin trajectory with the same endpoints and arrows reversed 5) The two trajectories together form a homoclinic orbit
Nondimensionalization: Pendulum
Pendulum Fixed points: (θ *, ν * ) = (kπ, 0), where k is any integer Focus: (0, 0), (π, 0) (the other fixed points coincide with either of them) (0, 0) -> τ = 0, Δ = 1 > 0 -> linear center! (π, 0) -> τ = 0, Δ = -1 < 0 -> saddle point!
Pendulum The system is reversible The system is conservative The energy function has a local minimum at (0, 0) The origin is then a nonlinear center
Pendulum The eigenvectors at the saddle fixed point are v 1 = (1, 1) and v 2 = (1, -1)
Pendulum Let us add the energy contours for different values of E Portrait is periodic in the θ-direction
Pendulum Physical interpretation: 1) Center -> pendulum at rest straight down (minimum energy E = -1) 2) Small orbits about the center -> small oscillations (librations) 3) If the energy increases, the amplitude of the oscillations increases 4) When E = 1 the motion asymptotically reaches the position of unstable equilibrium (pendulum straight up, heteroclinic orbit) 5) When E > 1 the pendulum whirls repeatedly over the top
Pendulum Cylindrical phase space Natural space for pendulum: one variable (θ) is periodic, the other (ν) is not Periodic whirling motions (E > 1) look periodic Saddle points indicate the same physical state Heteroclinic trajectories become homoclinic orbits
Pendulum Cylindrical phase space Plotting vertically the energy instead of the velocity: U-tube
Pendulum Cylindrical phase space Orbits are sections at constant height/energy The two arms correspond to the two senses of rotations Homoclinic orbits lie at E = 1, borderline between librations and rotations
All trajectories continuously lose altitude, except for the fixed points Pendulum Damping Centers -> stable spirals Saddle points -> saddle points
Pendulum Damping Change of energy along trajectory: Consequence: E decreases monotonically along trajectories, except at fixed points (where )
Pendulum Damping Physics: pendulum rotates over the top with decreasing energy, until it cannot complete the rotation and makes damped oscillations about equilibrium, where it eventually stops
Index theory Global information about the phase portrait, as opposed to the local information provided by linearization Questions: 1) Must a closed trajectory always encircle a fixed point? 2) If so, what types of fixed points are permitted? 3) What types of fixed points can coalesce in bifurcations? 4) Trajectories near higher-order fixed points? 5) Possibility of closed orbits? Index of a closed curve C: integer that measures the winding of the vector field on C Similarity with electrostatics: from the behavior of electric field on a surface one may deduce the total amount of charges inside the surface; here one gets info on possible fixed points
Index theory Vector field, simple (= non-self-intersecting) closed curve C, not passing through fixed points of the system At each point of C the vector field makes a well-defined angle with the positive x-axis
Index theory As x moves counterclockwise around C, the angle φ changes continuously (the vector field is smooth) -> when x comes back to the starting position φ has varied by a multiple of 2π [φ] C = net change in φ over one circuit Index of the closed curve C:
Example I What s the index? The vector field makes one complete rotation counterclockwise, so I C = + 1
Trick The index is the number of (counterclockwise) rotations that need to be done to go from 1 to 1 following the order of the numbers
Example II The vector field makes one complete rotation clockwise: I C = - 1
Example III The curve C is the unit circle: What is I C? [φ] C = -π + 2π -π = 0 I C = 0
Properties of the index 1) If C can be continuously deformed into C without passing through a fixed point, I C = I C. Proof: the index cannot vary continuously, but only by integer values, so it cannot be altered by a continuous change of C 2) If C does not enclose any fixed points, I C = 0. Proof: by squeezing C until it becomes a very small circle the index does not change because of 1), and it equals zero because all vectors on the tiny circle are essentially identical (as the field varies smoothly in space)
Properties of the index 3) Under time reversal (t -> -t), the index is the same. Proof: the time reversal changes the signs of the velocity vectors, so the angles change from φ to φ + π, hence [φ] C stays the same 4) If C is a trajectory of the system, I C = + 1
Index of a point The index of an isolated fixed point x * is the index of the vector field on any closed curve encircling x * and no other fixed point By property 1), the value of the index is the same on any curve C, since it can be continuously deformed onto any other What is the index of a stable node? The vector field makes one complete rotation counterclockwise, so I = + 1 The value is the same for unstable nodes as well, as the situation would be the same, only with reversed arrows (property 3)
Index of a point What is the index of a saddle point? The vector field makes one complete rotation clockwise: I = - 1 Spiral, centers, degenerate nodes and stars all have I = + 1, only saddle points have a different value
Index of a point Theorem: If a closed curve C surrounds n isolated fixed points, the index of the vector field on C equals the sum of the indices of the enclosed fixed points Proof: C can be deformed to the contour Γ of the figure; the contributions of the bridges cancel as each bridge is crossed in both directions so the net changes in the angle are equal and opposite
Index of a point Theorem: Any closed orbit in the phase plane must enclose fixed points whose indices sum to +1 Proof: If C is a closed orbit, I C = +1. From the previous theorem this is also the sum of the indices of the fixed points enclosed by C Consequence: any closed orbit encloses at least one fixed point (if there were none, the index on the curve would be 0, instead of + 1). If there is a unique fixed point, it cannot be a saddle (as in this case the index would be 1)
Example I Show that closed orbits are impossible for the rabbits versus sheep system The only orbits enclosing good fixed points (i.e. with index + 1) would have to cross the x/y axes, which contain trajectories of the system, and trajectories cannot cross!
Example II Show that the system has no closed orbits Solution: the system has no fixed points, so it cannot have closed orbits, since the latter have to enclose at least one fixed points
Exercises IV 1) Find and classify all fixed points, and sketch the nullclines and the phase portrait for 2) Find the fixed points, classify them, sketch the neighboring trajectories and try to fill in the phase portrait, for the systems